Solve Equations Using the Subtraction and Addition Properties of Equality

We are now ready to “get to the good stuff.” You have the basics down and are ready to begin one of the most important topics in algebra: solving equations. The applications are limitless and extend to all careers and fields. Also, the skills and techniques you learn here will help improve your critical thinking and problem-solving skills. This is a great benefit of studying mathematics and will be useful in your life in ways you may not see right now.

Solve Equations Using the Subtraction and Addition Properties of Equality

We began our work solving equations in previous chapters. It has been a while since we have seen an equation, so we will review some of the key concepts before we go any further.

We said that solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle.

In the earlier sections, we listed the steps to determine if a value is a solution. We restate them here.

Determine whether $y = \frac{3}{4}$ is a solution for $4 y + 3 = 8 y .$

Solution


$The text 'Substitute 3/4 for y.' is displayed on a white background. The word 'Substitute' is in dark blue, the fraction '3/4' is in red, and 'for y.' is in dark blue.$
Multiply.
Add.

Since $y = \frac{3}{4}$ results in a true equation, $\frac{3}{4}$ is a solution to the equation $4 y + 3 = 8 y .$

We introduced the Subtraction and Addition Properties of Equality in Solving Equations Using the Subtraction and Addition Properties of Equality. In that section, we modeled how these properties work and then applied them to solving equations with whole numbers. We used these properties again each time we introduced a new system of numbers. Let’s review those properties here.

When you add or subtract the same quantity from both sides of an equation, you still have equality.

We introduced the Subtraction Property of Equality earlier by modeling equations with envelopes and counters. Figure 1 models the equation $x + 3 = 8 .$

An envelope and three yellow counters are shown on the left side. On the right side are eight yellow counters.

The goal is to isolate the variable on one side of the equation. So we ‘took away’ $3$ from both sides of the equation and found the solution $x = 5 .$

Some people picture a balance scale, as in Figure 2, when they solve equations.

Three balance scales are shown. The top scale has one red weight on each side and is balanced. Beside it is “1 mass on each side equals balanced.” The next scale has two weights on each side and is balanced. Beside it is “2 masses on each side equals balanced.” The bottom scale has one weight on the left and two on the right. The right side is lower than the left. Beside the image is “1 mass on one side and 2 masses on the other equals unbalanced.”

The quantities on both sides of the equal sign in an equation are equal, or balanced. Just as with the balance scale, whatever you do to one side of the equation you must also do to the other to keep it balanced.

Let’s review how to use Subtraction and Addition Properties of Equality to solve equations. We need to isolate the variable on one side of the equation. And we check our solutions by substituting the value into the equation to make sure we have a true statement.

Solve: $x - 11 = −3 .$

Solution

To isolate $x,$ we undo the addition of $11$ by using the Subtraction Property of Equality.


We "undo" the subtraction of 11 by adding 11 to each side.
Simplify.
Check:
Substitute $x = 8$ .

Since $x = 8$ makes $x - 11 = −3$ a true statement, we know that it is a solution to the equation.

In the original equation in the previous example, $11$ was added to the $x$ , so we subtracted $11$ to ‘undo’ the addition. In the next example, we will need to ‘undo’ subtraction by using the Addition Property of Equality.

Solve: $m + 4 = −5 .$

Solution


Subtract 4 from each side to "undo" the addition.
Simplify.
Check:
Substitute $m = −9$ .

		The solution to $m - 4 = −5$ is $m = −1$ .

Now let’s review solving equations with fractions.

Solve: $n - \frac{3}{8} = \frac{1}{2} .$

Solution

		$A mathematical equation is displayed on a white background, which reads 'n - 3/8 = 1/2'. The variable 'n' is followed by a minus sign, then the fraction three-eighths, an equals sign, and finally the fraction one-half.$
Use the Addition Property of Equality.
Find the LCD to add the fractions on the right.
Simplify
Check:
$The text reads 'Substitute n = 7/8.' The word 'Substitute' and the variable 'n' are in a dark teal color, while the fraction '7/8' is in red. The equality sign '=' is also in teal.$	$A math equation reads '7/8 - 3/8 = ? 1/2'. The problem involves subtracting two fractions with the same denominator and comparing the result to one-half.$
Subtract.	$A mathematical equation showing the fractions 4/8 and 1/2 separated by a question mark, indicating a missing operator. The question asks what symbol should replace the question mark.$
Simplify.
The solution checks.

In Solve Equations with Decimals, we solved equations that contained decimals. We’ll review this next.

Solve $a - 3.7 = 4.3 .$

Solution


Use the Addition Property of Equality.
Add.
Check:
Substitute $a = 8$ .
Simplify.
The solution checks.

Solve Equations That Need to Be Simplified

In the examples up to this point, we have been able to isolate the variable with just one operation. Many of the equations we encounter in algebra will take more steps to solve. Usually, we will need to simplify one or both sides of an equation before using the Subtraction or Addition Properties of Equality. You should always simplify as much as possible before trying to isolate the variable.

Solve: $3 x - 7 - 2 x - 4 = 1 .$

Solution

The left side of the equation has an expression that we should simplify before trying to isolate the variable.


Rearrange the terms, using the Commutative Property of Addition.
Combine like terms.
Add 11 to both sides to isolate $x$ .
Simplify.
Check. Substitute $x = 12$ into the original equation.

The solution checks.

Solve: $3 (n - 4) - 2 n = −3 .$

Solution

The left side of the equation has an expression that we should simplify.


Distribute on the left.
Use the Commutative Property to rearrange terms.
Combine like terms.
Isolate n using the Addition Property of Equality.
Simplify.
Check. Substitute $n = 9$ into the original equation. The solution checks.

Solve: $2 (3 k - 1) - 5 k = −2 - 7 .$

Solution

Both sides of the equation have expressions that we should simplify before we isolate the variable.


Distribute on the left, subtract on the right.
Use the Commutative Property of Addition.
Combine like terms.
Undo subtraction by using the Addition Property of Equality.
Simplify.
Check. Let $k = −7 .$ The solution checks.

Translate an Equation and Solve

In previous chapters, we translated word sentences into equations. The first step is to look for the word (or words) that translate(s) to the equal sign. Table 9 reminds us of some of the words that translate to the equal sign.

Equals (=)
is	is equal to	is the same as	the result is	gives	was	will be

Let’s review the steps we used to translate a sentence into an equation.

Now we are ready to try an example.

Translate and solve: five more than $x$ is equal to $26 .$

Solution

Translate.
Subtract 5 from both sides.
Simplify.
Check: Is $26$ five more than $21$ ? The solution checks.

Translate and solve: The difference of $5 p$ and $4 p$ is $23 .$

Solution

Translate.
Simplify.
Check:
The solution checks.

Translate and Solve Applications

In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for.

The Robles family has two dogs, Buster and Chandler. Together, they weigh $71$ pounds.

Chandler weighs $28$ pounds. How much does Buster weigh?

Solution

A step-by-step guide demonstrating how to solve word problems using algebraic equations, exemplified by calculating a dog's weight.
Read the problem carefully.
Identify what you are asked to find, and choose a variable to represent it.	How much does Buster weigh? Let $b =$ Buster's weight
Write a sentence that gives the information to find it.	Buster's weight plus Chandler's weight equals 71 pounds.
We will restate the problem, and then include the given information.	Buster's weight plus 28 equals 71.
Translate the sentence into an equation, using the variable $b$ .
Solve the equation using good algebraic techniques.
Check the answer in the problem and make sure it makes sense.
Is 43 pounds a reasonable weight for a dog? Yes. Does Buster's weight plus Chandler's weight equal 71 pounds?
$43 + 28 \overset{?}{=} 71$
$71 = 71 ✓$
Write a complete sentence that answers the question, "How much does Buster weigh?"	Buster weighs 43 pounds

Shayla paid $$24,575$ for her new car. This was $$875$ less than the sticker price. What was the sticker price of the car?

Solution

Step-by-step solution for a word problem involving finding a car's sticker price, demonstrating variable assignment, equation translation, and verification.
What are you asked to find?	"What was the sticker price of the car?"
Assign a variable.	Let $s =$ the sticker price of the car.
Write a sentence that gives the information to find it.	$24,575 is $875 less than the sticker price $24,575 is $875 less than $s$
Translate into an equation.
Solve.
Check:
Is $875 less than $25,450 equal to $24,575?
$25,450 - 875 \overset{?}{=} 24,575$
$24,575 = 24,575 ✓$
Write a sentence that answers the question.	The sticker price was $25,450.

Key Concepts

Determine whether a number is a solution to an equation.
1. Substitute the number for the variable in the equation.
2. Simplify the expressions on both sides of the equation.
3. Determine whether the resulting equation is true.
If it is true, the number is a solution.
If it is not true, the number is not a solution.
Subtraction and Addition Properties of Equality
- Subtraction Property of Equality
  For all real numbers a, b, and c,
  if a = b then $a - c = b - c$ .
- Addition Property of Equality
  For all real numbers a, b, and c,
  if a = b then $a + c = b + c$ .
Translate a word sentence to an algebraic equation.
1. Locate the “equals” word(s). Translate to an equal sign.
2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
3. Translate the words to the right of the “equals” word(s) into an algebraic expression.
Problem-solving strategy
1. Read the problem. Make sure you understand all the words and ideas.
2. Identify what you are looking for.
3. Name what you are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Practice Makes Perfect

Solve Equations Using the Subtraction and Addition Properties of Equality

In the following exercises, determine whether the given value is a solution to the equation.

Is $y = \frac{1}{3}$ a solution of $4 y + 2 = 10 y ?$

Solution

yes

Is $x = \frac{3}{4}$ a solution of $5 x + 3 = 9 x ?$

Is $u = − \frac{1}{2}$ a solution of $8 u - 1 = 6 u ?$

Solution

Is $v = − \frac{1}{3}$ a solution of $9 v - 2 = 3 v ?$

In the following exercises, solve each equation.

$x + 7 = 12$

Solution

x = 5

$y + 5 = −6$

$b + \frac{1}{4} = \frac{3}{4}$

Solution

$b = \frac{1}{2}$

$a + \frac{2}{5} = \frac{4}{5}$

$p + 2.4 = −9.3$

Solution

p = −11.7

$m + 7.9 = 11.6$

$a - 3 = 7$

Solution

a = 10

$m - 8 = −20$

$x - \frac{1}{3} = 2$

Solution

$x = \frac{7}{3}$

$x - \frac{1}{5} = 4$

$y - 3.8 = 10$

Solution

y = 13.8

$y - 7.2 = 5$

$x - 15 = −42$

Solution

x = −27

$z + 5.2 = −8.5$

$q + \frac{3}{4} = \frac{1}{2}$

Solution

$q = - \frac{1}{4}$

$p - \frac{2}{5} = \frac{2}{3}$

$y - \frac{3}{4} = \frac{3}{5}$

Solution

$y = \frac{27}{20}$

Solve Equations that Need to be Simplified

In the following exercises, solve each equation.

$c + 3 - 10 = 18$

$m + 6 - 8 = 15$

Solution

m = 17

$9 x + 5 - 8 x + 14 = 20$

$6 x + 8 - 5 x + 16 = 32$

Solution

x = 8

$−6 x - 11 + 7 x - 5 = −16$

$−8 n - 17 + 9 n - 4 = −41$

Solution

n = −20

$3 (y - 5) - 2 y = −7$

$4 (y - 2) - 3 y = −6$

Solution

y = 2

$8 (u + 1.5) - 7 u = 4.9$

$5 (w + 2.2) - 4 w = 9.3$

Solution

w = −1.7

$−5 (y - 2) + 6 y = −7 + 4$

$−8 (x - 1) + 9 x = −3 + 9$

Solution

x = −2

$3 (5 n - 1) - 14 n + 9 = 1 - 2$

$2 (8 m + 3) - 15 m - 4 = 3 - 5$

Solution

m = −4

$- (j + 2) + 2 j - 1 = 5$

$- (k + 7) + 2 k + 8 = 7$

Solution

k = 6

$6 a - 5 (a - 2) + 9 = −11$

$8 c - 7 (c - 3) + 4 = −16$

Solution

c = −41

$8 (4 x + 5) - 5 (6 x) - x = 53$

$6 (9 y - 1) - 10 (5 y) - 3 y = 22$

Solution

y = 28

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve.

Five more than $x$ is equal to $21 .$

The sum of $x$ and $−5$ is $33 .$

Solution

x + (−5) = 33; x = 38

Ten less than $m$ is $−14 .$

Three less than $y$ is $−19 .$

Solution

y − 3 = −19; y = −16

The sum of $y$ and $−3$ is $40 .$

Eight more than $p$ is equal to $52 .$

Solution

p + 8 = 52; p = 44

The difference of $9 x$ and $8 x$ is $17 .$

The difference of $5 c$ and $4 c$ is $60 .$

Solution

5c − 4c = 60; c = 60

The difference of $n$ and $\frac{1}{6}$ is $\frac{1}{2} .$

The difference of $f$ and $\frac{1}{3}$ is $\frac{1}{12} .$

Solution

$f - \frac{1}{3} = \frac{1}{12}; f = \frac{5}{12}$

The sum of $−4 n$ and $5 n$ is $−32 .$

The sum of $−9 m$ and $10 m$ is $−25 .$

Solution

−9m + 10m = −25; m = −25

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Pilar drove from home to school and then to her aunt’s house, a total of $18$ miles. The distance from Pilar’s house to school is $7$ miles. What is the distance from school to her aunt’s house?

Jeff read a total of $54$ pages in his English and Psychology textbooks. He read $41$ pages in his English textbook. How many pages did he read in his Psychology textbook?

Solution

Let p equal the number of pages read in the Psychology book. 41 + p = 54. Jeff read 13 pages in his Psychology book.

Pablo’s father is $3$ years older than his mother. Pablo’s mother is $42$ years old. How old is his father?

Eva’s daughter is $5$ years younger than her son. Eva’s son is $12$ years old. How old is her daughter?

Solution

Let d equal the daughter’s age. d = 12 − 5. Eva’s daughter’s age is 7 years old.

Allie weighs $8$ pounds less than her twin sister Lorrie. Allie weighs $124$ pounds. How much does Lorrie weigh?

For a family birthday dinner, Celeste bought a turkey that weighed $5$ pounds less than the one she bought for Thanksgiving. The birthday dinner turkey weighed $16$ pounds. How much did the Thanksgiving turkey weigh?

Solution

21 pounds

The nurse reported that Tricia’s daughter had gained $4.2$ pounds since her last checkup and now weighs $31.6$ pounds. How much did Tricia’s daughter weigh at her last checkup?

Connor’s temperature was $0.7$ degrees higher this morning than it had been last night. His temperature this morning was $101.2$ degrees. What was his temperature last night?

Solution

100.5 degrees

Melissa’s math book cost $$22.85$ less than her art book cost. Her math book cost $$93.75 .$ How much did her art book cost?

Ron’s paycheck this week was $$17.43$ less than his paycheck last week. His paycheck this week was $$103.76 .$ How much was Ron’s paycheck last week?

Solution

$121.19

Everyday Math

Baking Kelsey needs $\frac{2}{3}$ cup of sugar for the cookie recipe she wants to make. She only has $\frac{1}{4}$ cup of sugar and will borrow the rest from her neighbor. Let $s$ equal the amount of sugar she will borrow. Solve the equation $\frac{1}{4} + s = \frac{2}{3}$ to find the amount of sugar she should ask to borrow.

Construction Miguel wants to drill a hole for a $\frac{5}{8} -inch$ screw. The screw should be $\frac{1}{12}$ inch larger than the hole. Let $d$ equal the size of the hole he should drill. Solve the equation $d + \frac{1}{12} = \frac{5}{8}$ to see what size the hole should be.

Solution

$d = \frac{13}{24}$

Writing Exercises

Is $−18$ a solution to the equation $3 x = 16 - 5 x ?$ How do you know?

Write a word sentence that translates the equation $y - 18 = 41$ and then make up an application that uses this equation in its solution.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A self-assessment chart with 'I can...' statements for solving equations, including using properties of equality, simplifying, translating, and solving applications. Options are 'Confidently,' 'With some help,' and 'No-I don't get it!'

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

solution of an equation

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.